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## Main Question or Discussion Point

[tex](a^{\frac{1}{2n}}a^{\frac{1}{2n}})^n=(a^{\frac{1}{2n}})^n(a^{\frac{1}{2n}})^n=(a^{\frac{1}{2}}) (a^{\frac{1}{2}})=a=(a^{\frac{1}{n}})^n[/tex]

all above is just done by using that the order of the factors that you multiply does not matter

we have proven that

[tex](a^{\frac{1}{p}}a^{\frac{1}{p}})^n=(a^{\frac{2}{p}})^n[/tex]

for any even number. Try saying that p is a odd number then n is a decimal number and taking the rooth with a number that is not whole does not make sense in a logixal approach of what one can really comprehend. For any even number of p what I did above makes sense. Could someone see an extention of this system. I want to make left side here:

[tex]\sqrt[m]{\underbrace{a^{\frac{1}{n}} \, \cdot \, a^{\frac{1}{n}} \, \cdot \, . . . \, \cdot a^{\frac{1}{n}} \, }_{\text{m times}} \, }=a^{\frac{m}{n}} [/tex]

to become right side by only using roths and powers of integers.

all above is just done by using that the order of the factors that you multiply does not matter

we have proven that

[tex](a^{\frac{1}{p}}a^{\frac{1}{p}})^n=(a^{\frac{2}{p}})^n[/tex]

for any even number. Try saying that p is a odd number then n is a decimal number and taking the rooth with a number that is not whole does not make sense in a logixal approach of what one can really comprehend. For any even number of p what I did above makes sense. Could someone see an extention of this system. I want to make left side here:

[tex]\sqrt[m]{\underbrace{a^{\frac{1}{n}} \, \cdot \, a^{\frac{1}{n}} \, \cdot \, . . . \, \cdot a^{\frac{1}{n}} \, }_{\text{m times}} \, }=a^{\frac{m}{n}} [/tex]

to become right side by only using roths and powers of integers.

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